A Pendulum on Mars. A certain simple pendulum has a period on the earth of 1.65 s. What is its period on the surface of Mars, where g = 3.71 m/s2?
2 Answers
-
T = 2π ∙ √(L/g)
where
T = period
L = length of pendulum
g = acceleration by gravity
On earth:
1.65 s = 2π ∙ √( L / (9.81 m/s²))
L = 0.6765 m
On mars:
T = 2π ∙ √( (0.6765 m) / (3.71 m/s²))
T = 2.68 s
-
T = 2 pi sqrt(L/g), so
L = g [ T / (2 pi) ]^2
For a constant L, we will have a constant gT^2.
Thus on Mars
3.71 T^2 = 9.8 (1.65 s)^2
T^2 = (9.8/3.71) (1.65 s)^2
T = (1.65 s) sqrt(9.8/3.71) = about 2.7 seconds but use calculator